# Find a real number $x$ such that $x^5 − x − 1 = 0$. [closed]

Find a real number $x$ such that $x^5 − x − 1 = 0$.

I have already proven that such a number exists, I now try to find which number it is.

• Iteration tells me its approximate value is $1.167303978$ (I iterated $x_{n+1}=\sqrt[5]{x_n+1}, x_1=1$) Jun 7, 2018 at 11:53
• A real polynomial with odd degree has at least a real root. Jun 7, 2018 at 11:55
• I'm voting to close this question because it's essentially answered already: you know there is a root. The comments tell you how to find a numerical approximation. There is no way to express that root with an expression involving roots and other algebra. Jun 7, 2018 at 11:58
• You can write the root as a nice-looking series
– user567668
Jun 7, 2018 at 12:03
• Do you know the notion of "Galois group"? That is the concept to use to answer whether or not the solution can be expressed in terms of radicals. Jun 7, 2018 at 13:15

You can use numerical methods such as Newtons method .

Define $h(x) = x^5-x-1$

$h'(x) = 5x^4-1$

$x_{n+1} = x_n-\dfrac{h(x_n)}{h'(x_n)}$

$x_1 = x_0 - \dfrac{h(x_0)}{h'(x_0)}$

With $x_0 = 1$ we get the following;

$\begin{pmatrix}n&&&&x_n\\0&&&&1\\1&&&&1.25\\2&&&&1.178459\\3&&&&1.167537\\4&&&&1.167304\\\vdots\end{pmatrix}$

• Right, so this defines a Cauchy sequence $x_n$ that converges to a real number that is the solution of $x^5-x-1$. If we take equivalence class of this Cauchy sequence, we have now successfully constructed the real number we were looking for. Jun 7, 2018 at 12:08
• yes , thats correct. Jun 7, 2018 at 12:09

You can use an iteration to find its approximate value:

Rearrange for:$$x^5=x+1\to x=\sqrt[5]{x+1}$$ Then iterate: $$x_{n+1}=\sqrt[5]{x_n+1}, x_1=1$$

Calculator accuracy gives me $x\approx1.167303978$